Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 m/s into a cyl- indrical tank, the radius of whose base is 60 cm, By how much will the level of water rise in 30 min?

Options

A. 3m
B. 3.5 m
C. 4 m
D. 4.5 m

Answer

To solve the problem, we need to calculate how much the water level in the cylindrical tank rises when water flows out of a circular pipe. Here are the steps to find the solution:

### Step 1: Calculate the radius of the pipe
The internal diameter of the pipe is given as 2 cm. To find the radius, we divide the diameter by 2.

\[ \text{Radius of the pipe} = \frac{\text{Diameter}}{2} = \frac{2 \text{ cm}}{2} = 1 \text{ cm} \]

### Step 2: Convert the radius to meters
Since the flow rate is given in meters per second, we should convert the radius from centimeters to meters.

\[ \text{Radius of the pipe in meters} = 1 \text{ cm} = 0.01 \text{ m} \]

### Step 3: Calculate the cross-sectional area of the pipe
The cross-sectional area \(A\) of the circular pipe can be calculated using the formula for the area of a circle:

\[ A = \pi r^2 \]

Substituting the radius we found:

\[ A = \pi (0.01 \text{ m})^2 = \pi (0.0001 \text{ m}^2) \approx 0.000314 \text{ m}^2 \]

### Step 4: Calculate the volume of water flowing out in 30 minutes
The flow rate of water is given as 6 m/s. To find the total volume of water that flows out in 30 minutes, we first convert 30 minutes into seconds:

\[ 30 \text{ minutes} = 30 \times 60 = 1800 \text{ seconds} \]

Now, we can calculate the total volume \(V\) of water flowing out:

\[ V = \text{Flow rate} \times \text{Time} = 6 \text{ m/s} \times 1800 \text{ s} = 10800 \text{ m}^3 \]

### Step 5: Calculate the volume of water in cubic meters
Since we need the volume in cubic meters, we have already calculated it as 10800 m³.

### Step 6: Calculate the rise in water level in the cylindrical tank
The radius of the base of the cylindrical tank is given as 60 cm, which is 0.6 m. The area of the base \(A_t\) of the cylindrical tank can be calculated as:

\[ A_t = \pi r^2 = \pi (0.6 \text{ m})^2 = \pi (0.36 \text{ m}^2) \approx 1.131 \text{ m}^2 \]

Now, we can find the rise in water level \(h\) using the formula for volume:

\[ V = A_t \times h \implies h = \frac{V}{A_t} \]

Substituting the values we have:

\[ h = \frac{10800 \text{ m}^3}{1.131 \text{ m}^2} \approx 9555.78 \text{ m} \]

### Step 7: Final Result
The rise in the water level in the cylindrical tank is approximately 9555.78 m, which seems incorrect based on the context of the problem. Let's check calculations again.

### Correct Calculation of Volume Flow
The correct volume flow from the pipe should be calculated as follows:

\[ \text{Volume flow rate} = \text{Area of pipe} \times \text{Velocity} = 0.000314 \text{ m}^2 \times 6 \text{ m/s} \approx 0.001884 \text{ m}^3/s \]

Now, for 30 minutes:

\[ V = 0.001884 \text{ m}^3/s \times 1800 \text{ s} \approx 3.3912 \text{ m}^3 \]

Now, we can calculate the rise in water level again:

\[ h = \frac{3.3912 \text{ m}^3}{1.131 \text{ m}^2} \approx 2.997 \text{ m} \]

### Conclusion
The level of water will rise approximately 3 m in the cylindrical tank in 30 minutes.

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